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Recognizing a product of two 3d rotations (matrices)

  1. Jun 14, 2008 #1
    Hi, I have a problem identifying some 3d rotation matrices. Actually I don't know if the result can be brought on the desired form, however it would make sense from a physics point of view. My two questions are given at the bottom.

    [tex]\mathbf{s}=\left(
    \begin{array}{c}
    s_{x} \\ s_{y} \\ s_{z}
    \end{array} \right),\; \mathbf{S}=\left(
    \begin{array}{c}
    S_{x} \\ S_{y} \\ S_{z}
    \end{array} \right)[/tex]

    [tex]H=\left[\left(
    \begin{array}{ccc}
    1-\left(1-C_{z}^{2}+C_{x}^{2}\right)\gamma^{2}& 0 & -2\gamma\left(1-C_{x}C_{z}\gamma\right) \\
    0 & 1-\left(1+C_{z}^{2}+C_{x}^{2}\right)\gamma^{2} & 0 \\
    2\gamma\left(1-C_{x}C_{z}\gamma\right)& 0 &
    1-\left(1+C_{z}^{2}-C_{x}^{2}\right)\gamma^{2}
    \end{array} \right) \mathbf{s}\right]\cdot\mathbf{S}
    [/tex]

    (this describes Kondo effect in a quantum dot with spin-orbit interaction)

    The goal is to bring H on a the form [tex]H = \left(\mathbf{A}\mathbf{s}\right)\cdot\left(\mathbf{B}\mathbf{S}\right)[/tex] where A and B are matrices describing rotations.


    For Cx=Cz=0 :


    [tex]H=\left[\left(1-\gamma^{2}\right)\underbrace{\left(
    \begin{array}{ccc}
    1 & 0 & \theta \\
    0 & 1 & 0 \\
    -\theta & 0 & 1
    \end{array} \right)}_{R_{y}(\theta)+O(\theta^{2})}
    \mathbf{s}\right]\cdot\mathbf{S} \;\; \approx \;\; \left(1-\gamma^{2}\right)\left(R_{y}(\theta) \mathbf{s}
    \right)\cdot \mathbf{S} \;, \qquad \theta=\frac{-2\gamma}{1-\gamma^{2}}
    [/tex]

    That is, H is a vector product between spin S and a spin s that has been rotated around the y-axis with angle theta.

    For Cx and Cz different from 0 :

    In this case the matrix cannot be direct identified as a rotation around the x,y or z axis. The questions are now:

    1) can it be a rotation around another axis?
    2) Is it possible to write it as [tex]H = \left(\mathbf{A}\mathbf{s}\right)\cdot\left(\mathbf{B}\mathbf{S}\right)[/tex] where A and B are matrices describing rotations?

    I would be glad if anybody has an idea about how to deal with this.
     
  2. jcsd
  3. Jun 14, 2008 #2
    You essentially have written that [tex]H = \mathbf{s}_1^T \mathbf{D} \mathbf{s}_2[/tex], and the question is whether you can write [tex]\mathbf{D} = \mathbf{A} \mathbf{B}[/tex], where A and B are rotation matrices. Now, rotation matrices form a group, hence [tex]\mathbf{D}[/tex] should also be a rotation matrix. Is [tex]\mathbf{D}[/tex] orthogonal? I'm assuming it is.

    Anyway, there isn't a unique way of writing D as a product of two rotations. One is to take A = D and B = identity. Maybe I've misunderstood.
     
  4. Jun 16, 2008 #3
    One of the real eigenvectors of the matrix is the y unit vector, so if it is a rotation matrix it can only possibly be a rotation about the y-axis. If it is a rotation matrix, the corresponding eigenvalue
    [tex] 1-\left(1+C_{z}^{2}+C_{x}^{2}\right)\gamma^{2}[/tex],
    must be equal to 1. This can only happen if \gamma = 0. So except in this presumably trivial case, no, it is not a rotation matrix.

    Dave
     
    Last edited: Jun 16, 2008
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